斐波拉契数列

斐波拉契数列三种解法：

#include<iostream>
using namespace std;
 
int f[100];
int fib(int n){
    if(n==0) return 0;
    if(n==1) return 1;
    return fib(n-1)+fib(n-2);
}
 
 
 
int main(){
    int n;
    while(cin >> n){
        for(int i = 0; i< n;i++)
        cout << fib(i) << " ";
        cout << endl;
    }
 
 
return 0;
}

#include<iostream>
using namespace std;
 
int f[100];
int fib(int n){
    for(int i = 0;i < n; i++){
        if(i == 0 || i == 1) f[i] = i;
        else f[i] = f[i-1] + f[i-2];
    }
    for (int i =0; i< n-1 ;i++)
        cout << f[i] << " ";
    cout << f[n-1] << endl;
}
 
int main(){
    int n;
    while(cin >> n){
        fib(n);
    }
 
return 0;
}

#include<iostream>
#include<string>
using namespace std;
int fib(int n){
    if(n<1)
        return 0;
    if(n==1||n==2)
        return 1;
    int pre = 1;
    int res[n+1] = {0,1,1};
    int temp = 0;
    for(int i = 2;i<n;i++){
        //第n个数
        temp = res[i];
        res[i+1] = pre+res[i];
        pre = temp;
    }
    for(int i = 0;i<n;i++)
        cout << res[i]<<" ";
    cout << endl;
    return res[n-1];
}
 
int main(){
    cout << fib(12);
return 0;
}